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07_SEM_Chap5text - + 2nd 1. Step Input order G(s) A sudden...

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Unformatted text preview: + 2nd 1. Step Input order G(s) A sudden change in a process variable can be approximated by a step change of magnitude, M: In analyzing process dynamic and process control systems, it is important to know how the process responds to changes in the important to know how the process responds to changes in the process inputs. 1st-order response A number of standard types of input changes are widely used for number of standard of input changes are widely used for two reasons: 1. They are representative of the types of changes that occur in real operating processes. Chapter 5 Chapter 5 Dynamic Behavior: 1st 2. They are easy to analyze mathematically. Us ⎧0 t<0 ⎨ ⎩M t ≥ 0 (5-4) The step change occurs at an arbitrary time denoted as t = 0. step change occurs at an arbitrary time denoted as 0. • Special Case: If M = 1, we have a “unit step change”. We give it the symbol, give it the symbol, S(t). • Example of a step change: A reactor feedstock is suddenly switched from one supply to another, causing sudden changes in feed concentration, flow, etc. 1 2 Example: We can approximate a drifting disturbance by a ramp input: and Q ( t ) = 8000 + 2000S ( t ) , S (t ) Q′ ( t ) = Q − Q = 2000 S ( t ) , 1st-order response U R (t ) Q = 8000 kcal/hr unit step 2. Ramp Input • Industrial processes often experience “drifting disturbances”, that is, relatively slow changes up or down for some period of time. • The rate of change is approximately constant. The deviation variable is UR(t). 3 Chapter 5 Chapter 5 The heat input to the stirred-tank heating system in Chapter 2 is heat input to the stirred heating system in Chapter is suddenly changed from 8000 to 10,000 kcal/hr by changing the electrical signal to the heater. Thus, ⎧0 t < 0 ⎨ ⎩a it t ≥ 0 (5-7) Examples of ramp changes: of ramp changes: 1. Ramp a setpoint to a new value. (Why not make a step change?) change?) 2. Feed composition, heat exchanger fouling, catalyst activity, ambient temperature. 3. Rectangular Pulse Input It represents a brief, sudden change in a process variable: represents brief sudden change in process variable: 4 U RP ( t ) ⎧0 ⎪ ⎨h ⎪0 ⎩ for t < 0 for 0 ≤ t < tw (5-9) for t ≥ tw Chapter 5 Chapter 5 XRP h 0 tw Time, t Fig. 5.2 Three important examples of deterministic inputs. Examples: 1. Reactor feed is shut off for one hour. 2. The fuel gas supply to a furnace is briefly interrupted. 5 6 5. Impulse Input 4. Sinusoidal Input (later = Chap. 14, Sinusoidal Ouput) Processes are also subject to periodic, or cyclic, disturbances. They can be approximated by a sinusoidal disturbance: ⎧0 for t < 0 (5-14) ⎨ ⎩ A sin (ωt ) for t ≥ 0 2nd-order response may have increased amplitude at some frequencies! where: Here, U I ( t ) = δ ( t ) . It represents a short, transient disturbance. Examples of approximate impulses: approximate impulses: Chapter 5 Chapter 5 U sin ( t ) • • A = amplitude, ω = angular frequency, 1/time amplitude, angular frequency, 1/time = (radians)/time not 1. Electrical noise spike in a thermo-couple reading. 2. Injection of a tracer dye. • Useful for analysis since the time response to an impulse input is the inverse Laplace transform of the TF Thus input is the inverse Laplace transform of the TF. Thus, u (t ) U (s) (cycles)/time Examples of “cycles”: 1. 24 hour variations in cooling water temperature. 2. 60-Hz electrical noise (in the USA) Here, 7 G (s) y (t ) Y (s) Y ( s ) = G ( s )U ( s ) (1) 8 First-Order System Th The corresponding time domain express is: 0 Chapter 5 where: g ( t ) L−1 ⎡G ( s ) ⎤ ⎣ ⎦ Y (s) (2) Chapter 5 t y ( t ) = ∫ g ( t − τ ) u ( τ ) dτ The standard form for a first-order TF is: fi TF (3) Suppose u ( t ) = δ ( t ) . Then it can be shown that: y (t ) = g (t ) (4) Consequently, g(t) is called the “impulse response function”. where: U (s) K τ = K τs + 1 (5-16) steady-state gain time constant Consider the response of this system to a step of magnitude, M: U ( t ) = M for t ≥ 0 ⇒ U (s) = M s Substitute into (5-16) and rearrange, Y (s) = KM s ( τs + 1) (5-17) 9 10 Integrating Process Take L-1 (cf. Table 3.1) = memorize it ! , ( y ( t ) = KM 1 − e−t / τ steady-state value of y(t). From (5-18), y∞ = KM . 1.0 y y∞ 0.5 0 0 Not all processes have a steady-state gain. For example, an “integrating process” or “integrator” has the transfer function: (5-18) 1 2 3 t τ 4 5 t 0 τ 2τ 3τ 4τ 5τ Chapter 5 Chapter 5 Let y∞ ) y y∞ ___ 0 0.632 0.865 0.950 0.982 0.993* nearly complete Note: Larger τ means a slower response. 11 Y (s) U (s) = K s ( K = constant ) Consider a step change of magnitude M. Then U(s) = M/s and, -1 KM L Y ( s ) = 2 ⇒ y ( t ) = ( KM ) t s Thus, y(t) is unbounded and a new steady-state value does not exist, i.e. there is no Final Value ! 12 Common Physical Example: Second-Order Systems Consider a liquid storage tank with a pump on the exit line: li li • Standard form: Y (s) - Chapter 5 Chapter 5 U (s) Assume: 1. Constant cross-sectional area, A. - - 2. q ≠ f ( h ) this is not the check quiz story problem ! dh Mass balance: A = qi − q ⇒ 0 = qi − q s.s. equation dt Eq. (1) – Eq. (2), take L, assume steady state initially, 1 ⎡Qi′ ( s ) − Q′ ( s ) ⎤ H ′( s) = ⎦ H ′( s ) 1 As ⎣ = Qi′ ( s ) As For Q′ ( s ) = 0 (constant q), = K 22 τ s + 2ζτs + 1 (5-40) which has three model parameters: K steady-state gain τ "time constant" >0 [=] time constant" >0 time ζ damping coefficient (dimensionless) 1⎞ ⎛ • Equivalent form: ⎜ ωn natural frequency = ⎟ τ⎠ ⎝ Y (s) U (s) = 2 K ωn 2 s 2 + 2ζ ω n s + ω n 13 14 • The type of behavior that occurs depends on the numerical value of damping coefficient, ζ : It is convenient to consider three types of stable behavior: Type of Response Roots of Charact. Polynomial ζ >1 Overdamped Negative real and ≠ ζ =1 Critically damped Negative real and = Underdamped Complex conjugates negative real parts 0 ≤ ζ <1 Chapter 5 Chapter 5 Damping Coefficient Note: this is for this is for dimensionless time, “Speed” of approach to final value depends on the final value depends on the time constant (tau) – and only “somewhat” on zeta. • Note: The characteristic polynomial is the denominator of the The characteristic polynomial the denominator of the transfer function: τ 2 s 2 + 2ζτs + 1 • What about ζ < 0 ? It results in an unstable system 15 16 1. Responses exhibiting oscillation and overshoot (y/KM > 1) are obtained only for values of ζ less than one. Again - dimensionless time, of approach “Speed” of approach to final value depends on the time constant (tau) – and only “somewhat” on zeta. Chapter 5 Chapter 5 Several general remarks can be made concerning the responses show in previous two Figs. 5.8 and 5.9: 2. Large values of ζ yield a sluggish (slow) response for fixed τ . 3. The fastest response without overshoot is obtained for the fastest response without overshoot obtained for the critically damped case for fixedτ . ( ζ = 1) if the two time constants are identical. 17 18 1. Rise Time: tr is the time the process output takes to first reach the new steady-state value. Chapter 5 Chapter 5 2. Time to First Peak: tp is the time required for the output to reach its first maximum value. 3. Settling Time: ts is defined as the time required for the process output to reach and remain inside a band whose width is equal to ±5% of the total change in y. The term 95% Th 95% response time sometimes is used to refer to this case. Also, values of ±1% sometimes are used. 4. Overshoot: OS = a/b (% overshoot is 100a/b). 5. Decay Ratio: DR = c/a (where c is the height of the second Ratio DR the height of the second peak). 6. Period of Oscillation: P is the time between two successive peaks or two successive valleys of the response. 19 20 Second Order Step Change a. Block Notation: overshoot ζ >1 b. time of first maximum ⎛ πτ tp = ⎜ ⎜ 1− ζ 2 ⎝ c. ⎞ ⎟ ⎟ ⎠ decay ratio (successive maxima – not min.) ratio (successive maxima min ⎛ −2πζ c = exp ⎜ ⎜ 1− ζ 2 a ⎝ ⎞ a2 ⎟= 2 ⎟b ⎠ d. period of oscillation ⎛ 2πτ P=⎜ ⎜ 1− ζ 2 ⎝ ⎞ ⎟ ⎟ ⎠ overdamped Composed of two first order subsystems (G1 and G2) Chapter 5 Chapter 5 ⎛ −πζ ⎞ a = exp ⎜ ⎟ depends only on ζ , the damping coefficient 2⎟ ⎜ b ⎝ 1− ζ ⎠ G ( s )= K General 2nd-order K ζ= τ1 + τ 2 2 τ1τ 2 K D( s) Roots of D( s ) = 0 are poles of G ( s ). G (s )= 2 "poles" are at {s1 , s2 } = 21 τ = τ 1τ 2 τ 1τ 2 s 2 +(τ 1 + τ 2 )s +1 τ 2 s 2 + 2ζτ s + 1 −ζ ± ζ 2 − 1 τ = ζ <1 ζ =1 underdamped critically damped 22 ...
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This note was uploaded on 03/01/2012 for the course CHE 361 taught by Professor Staff during the Winter '08 term at Oregon State.

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